Table of Contents Module 1 11 Order of Operations 16 Signed Numbers 1 Factorization of Integers 17 Further Signed Numbers 13 Fractions 18 Power Laws 14 Fractions and Decimals 19 Introduction to Algebra 15 Percentages 110 Further Algebra Module 1 Factors of Algebraic Epressions 6 Factorizing Algebraic Epressions Solving Equations in One Variable 7 Logarithms and Eponentials 3 Algebraic Fractions 8 Introduction to Trigonometr 4 Introduction to Inequalities 9 Introduction to the Cartesian Plane 5 Arithmetic with Surds Module 3 31 Functions 36 Arithmetic & Geometric Progressions 3 Graphs 37 Continuit & Limits 33 Trigonometr 38 Introduction to Differentiation 34 Further Trigonometr 39 Further Differentiation 35 Simultaneous Equations 310 Differentiating Special Functions Module 4 41 More Differentiation 47 Polnomials 4 Introduction to Integration 48 Properties of Trigonometric Functions 43 Integrating Special Functions 49 44 Applications of Integration 410 45 Binomial Coefficients 411 46 Sigma Notation 41 Induction 1
Test Three This is a self-diagnostic test Ever pair of questions relates to a worksheet in a series available in the MUMS the WORD series For eample question 5 relates to worksheet 35 Simultaneous Equations If ou score 100% on this test and test 4 then we feel ou are adequatel prepared for our first ear mathematics course For those of ou who had trouble with a few of the questions, we recommend working through the appproriate worksheets and associated computer aided learning packages in this series 1 (a) If f() = +, what is f(3)? (b) If g() =, and f() = + 1, what is g f()? (a) What is the domain of g() in the following graph? g() 4 4 (b) At what points on the graph above does g() = 0? 3 (a) If an angle is 60, how man radians is it? (b) For the drawn below, what is the angle φ? φ 3 4 (a) What is sin 7π? (Without a calculator) 4 (b) Find the area of the triangle drawn below π 3
5 (a) Given: = + 5 = k + 4 For what value(s) of k will the sstem have no solutions? (b) Solve the sstem: = 3u + 6 3 = 5u + 6 (a) If, 7 are the first two terms of an arithmetic progression, what is the 10th term? (b) What is the sum of the following infinite geometric series? 1 + 9 10 + 81 100 + 7 (a) Find the limit of (b) Is the function f() = 5 lim n 5 5 { +6 0 0 = 0 continuous at = 0? Wh? 8 (a) Find the derivative of f() = 3 + 3 + 3 (b) What are the stationar points of the function g() = 3 + 3? 9 (a) What sort of turning points does the function f() = 6 3 have? (b) When does the concavit change for the function h() = 3 + + 5 +? 10 (a) Differentiate = sin(5 + 3) (b) Differentiate = 3e 3
Worksheet 31 Functions Section 1 Definitions What is a function? A function can be thought of as a machine It accepts an input, applies a rule to it and then produces an output Diagrammaticall, we might view the process like: input rule output Eample 1 : Rule f: take the input, and multipl it b 5 If we appl rule f to the input 4, we get 5 4 = 0 4 5 4 0 What is the output when we appl rule f to the input? 5 5 As mentioned in other worksheets we look for shorthand was of working with things The shorthand wa of writing appl rule f to input 4 is to write f(4) We sa this as f of 4 So f(4) = 0 and f() = 5 We sa the second item as f of When we appl rule f to input our output gives us a shorthand wa of writing the actual rule Eample : We define rule G: take the input squared, and then add 5 Appl rule G to the inputs 1, 1, a + 1 and G( 1) = ( 1) + 5 = 6 G(1) = (1) + 5 = 6 G(a + 1) = (a + 1) + 5 = a + a + 6 G() = + 1 There are several different was of representing functions The most common was are 4
1 As a table of values As a graph 3 As an algebraic epression Here are some eamples of the was of representing a function Eample 3 : input 0 1 3 4 5 6 output 5 6 7 8 9 10 11 This is a table of values, and the rule isn t given eplicitl in this case However, we might be able to guess an appropriate rule In this case it appears to be to take the input and add 5 With a table of values, the rule will not usuall be given, and it ma not be obvious what the rule is But the table of values still represents a function Let s give the rule that converts an input into an output a name, sa f Then the function that is associated with this table is f() = + 5 Eample 4 : This graph also represents a function, and from the graph we can learn things about the function We will talk more about graphs in the net worksheet Eample 5 : Consider f() = 5 + Here we use the shorthand notation for the function rule, and the input is alwas some variable which is often or t The output is given b what the rule does to the variable, in this case Rule f in this case sas take the input, square it, multipl the result b 5, and then add This is the output 5
Eercises: 1 For each part, find a function which describes the table of values (a) (b) (c) 0 1 3 f() 4 3 1 4 5 6 f() 5 17 6 37 0 4 5 f() 1 5 9 11 Complete the table of values for the function f() = 3 3 1 3 5 7 f() Section Substitution When a function is represented algebraicall, we are given the rule as it applies to some variable This is called functional notation To compute the rule applied to an input we simpl replace the variable with the input Eample 1 : Given: f() = then f(5) = (5) = 5 f( 1) = ( 1) = 1 f(a + b) = (a + b) = a + ab + b f() = () = 4 Eample : Given: g(t) = 5t 3 then g(1) = g(0) = 3 g(10) = 47 g( ) = 5 3 6
Eample 3 : Given: h() = 1 then h(1) = 1 1 = 1 h(3) = 1 3 In eample 3, h() = 1 So h(0) doesn t make sense since we can t divide b zero When the function doesn t make sense for a particular input value, we sa that the function is not defined for that input value Eample 4 : Given: f() = 3 + + then f(1) = 3 (1) + 1 + = 7 f(0) = f( 1) = 3 f(3) = 7 + 6 + Eercises: 1 Given f() =, find (a) f(3) (b) f( ) (c) f( 1) Given f() = + 3, find (a) f(4) (b) f( 3) (c) f( + ) 3 Given f() = ( + ) + 3, find (a) f() (b) f( ) (c) f( 4) (d) f( + 1) 7
Section 3 Composition of Functions This section deals with a thing called composition of functions As a picture, composition looks like this: input 1 rule 1 output 1 = input rule output It is like having two machines, one after the other The result from one machine forms the input to the net machine We could also write it like this rule f f() rule g g(f()) So g(f()) is a composite function which ma also be written g f() The circle ma be taken to mean follows Note: It is important to realize that g f() f g() The order in which the functions are applied is important It is equall, if not more important, to realize that f g() f() g() Eample 1 : Take g() = and f() = 1 + Then g f(1) = g(f(1)) = g() = 4 g f(3) = g(f(3)) = g(4) = 16 f g(1) = f(g(1)) = f(1) = f g(3) = f(g(3)) = f(9) = 10 g f() = g(f()) = g( + 1) = ( + 1) f g() = f(g()) = f( ) = + 1 When dealing with compositions, if ou write it all out longhand as in the above eamples, ou shouldn t get too confused It s when ou tr and do too much in our head that ou get the computation around the wrong wa 8
Eercises: 1 Given f() = + and g() =, find (a) f(3) (b) g(5) (c) g f(3) (d) g f( 1) (e) f g(3) (f) f g( 4) Given f() = 1 and g() = 3, find (a) g f(1) (b) g f(t) (c) g f(4) (d) f g( + 1) (e) f g( + ) Section 4 Functions from words Functions are useful for determining the answer to man problems that occur in real-life situations You ma be required to take a problem that is given to ou in words and come up with the function describing the information given There is no hard and fast method of dealing with problems, although there are some general hints 1 Read the information carefull, and translate as much as possible into mathematical epressions Tr out a few inputs to get a feel for the rule before writing it down with a variable Eample 1 : You have to create a rectangular paddock and ou onl have 1000 metres of fencing You can choose the width of the paddock Find a function that takes as input the width of the paddock and gives as an output the area of the paddock enclosed The perimeter of the paddock is 1000m If the length is l and the width w then the perimeter p in smbols is p = l + w The area A is A = lw, which we will tr to write in terms of w, rather than w and l From l + w = 1000, we see that Then the area is l = 1000 w l = 1 (1000 w) = 500 w A = lw = (500 w)w = 500w w 9
We can create the following table: The required function is w l A 300 500 300 = 00 300 00 00 500 00 = 300 00 300 100 500 100 = 400 100 400 500 (500 ) A = f() = (500 ) where A is an obvious smbol to represent area In later worksheets on differentiation we will learn a technique that will allow us to find the optimum width so that fencing of the paddock will ield the maimum area Eample : A ferr carries an average of 300 people a da The fare is $ 10 The UTA research shows that 50 etra people will travel per da for ever 10cent fare reduction Work out the function that has the number of fare reductions as input, and as output the total amount of mone collected b the UTA each da Reductions Fare Number of People Mone collected 0 10 300 300 10 100 300 + 50 (300 + 50) 100 1 000 300 + 1 50 (300 + 1 50) 000 10 010 300 + 50 (300 + 50) (10 010) So the function in terms of the number of reductions is (300 + 50) (10 010) In the worksheet on differentiation we will learn how to maimize the mone taken in b the UTA Eercises: 1 A truck weighs 1500 kg and it is to be loaded with cartons each weighing 5 kg Work out the function which has the number of cartons as input and the total weight of the truck as output A photocopier service costs $40 plus 5 cents for ever cop made Work out the function which has the number of copies made as the input and the total cost as output 10
Eercises for Worksheet 31 1 (a) A mother records the height of her son over the first 8 months, and measurements she made are shown in this table: Input (age in months) 0 4 6 8 Output (height in cm) 50 54 58 6 66 Epress the output as a function of the input (b) Consider the pattern of triangles shown: If the input is the number of horizontal rows in the pramid, and the output is the number of triangles, describe the relationship between the input and the output (c) Evaluate, given that = a 3 3 a, when a = 73 (d) Evaluate a, given that v = u + as, when v = 10 m/s, u = 9 m/s, and s = m (e) The formula for converting degrees Fahrenheit (F ) to Celsius (C) is given b C = 5 (F 3) Evaluate C when F = 100 9 (f) If f() = 3 + 1, find f( ) (g) If g() = 3+ 1, find g( 1 ) (h) If f() = + 3, find f( + h) (i) If f() = 3 + 4, find f(+h) f() h (j) For f() given in the previous question, evaluate f(+h) f() h when = and h = 0001 (a) If f() = and g() = 1, find i f g() ii g f() (b) If f() = 3 and g() = 3, find i f g() ii g f() (c) If f() =, g() = + 1, and h() =, find f g h() (d) If f() =, find i f() ii f( + h) iii f() 11
iv f( + 1) (e) If f() = 1 +1, find i f( 1 ) ii f(3 + ) iii f( ) 3 (a) Here is a rule for a function: take the input, multipl it b 3, then add 4, then square the result i Epress the output as a function of the input ii Evaluate the output when the input equals - (b) Here is another rule for a function: take the input, subtract, take the square root of the result, then add 5 i Epress the output as a function of the input ii Evaluate the output when the input equals (c) There are 7 times as man cars as motorccles in Australia If C represents the number of cars, and M the number of motorccles, write an equation describing the relationship between M and C 1
Worksheet 3 Graphs Section 1 Range & Domain In the last worksheet we mentioned that functions can be represented as graphs Graphs have alread been referred to in worksheet 10 when we looked at graphing a straight line The graph of a function is the collection of all points (, ) that satisf a given function From looking at graphs we can learn a lot of information about the function it represents From a graph, we can make estimates about the value of the function at certain inputs; we can see where maimum and minimum values of the function are; we can see how rapidl the function is increasing or decreasing Two important pieces of information we can read off a graph are the range and domain of the function The range of a function is all the values that the function takes So if is a function of, the range is all the -values that can be taken The range ma be written in one of several was - tpicall as an interval, or using inequalit signs Eample 1 : The arrows on the graph of f() indicate that it keeps going upwards The range of f() can be written as That is, can be greater than or equal to In interval notation this is written as [, ) Note: The square bracket indicates that is included in the range indicates all numbers up to but not including the end part So The round bracket (, ) is the same as > and [1, 3) is the same as 1 < 3 13
Eample : 1 g() The open circle on a graph means the same as an open circle on a number line: all numbers up to but not including the point Thus the range of g() is 1 < 1, or in interval notation [ 1, 1) 1 The domain of a function is all the inputs that make sense In other words, for a function f() it is all the -values for which the function is defined The domain of a function is normall written in the same notation as the range ie either using inequalities or interval notation Eample 3 : f() = 1 The domain is > 0 and < 0 Or equivalentl, (, 0) and (0, ) f() does not make sense for = 0, so it is not included in the domain The range of this function is > 0 and < 0 Or, equivalentl, (, 0) (0, ) Eample 4 : The domain of this graph is 3 1 3 3 0 and 1 3, or in interval notation [ 3, 0] and [1, 3] 14
Eercises: 1 Find the range and domain of the following graphs (a) (b) 3 (c) 8 Section Intercepts and Reading Graphs The value of a function when the input is zero (ie the value when = 0) is called the -intercept This is where the function crosses the -ais The input which gives the function a value of zero (ie the values that give = 0) are called the -intercepts The are called the -intercepts because this is where the plot of the function crosses the -ais To find the -intercept, we simpl substitute = 0 into the function The output is the -intercept To find the -intercept, we let = 0 and then solve the equation for This ma not alwas be a simple procedure The intercepts give us the beginning of a picture of the function and will help us to represent the function graphicall Eample 1 : 1 1 The -intercept is +1 The -intercept is 1 ie when = 0, = 1 and when = 0, = 1 15
Eample : B A C f() f() has a zero value at A and C; B marks the -intercept There are man possible formulae for the graph, and one possibilit is a parabola Eample 3 : Let = 3 + 6 Then at = 0, = 3 0 + 6 = 6 So the -intercept is 6 Now set = 0 to find the -intercept 0 = 3 + 6 = So the -intercept is Since = 3+6 is the equation for a straight line, we can now draw the graph The intercepts of a straight line give us enough information to draw the graph (unless both intercepts are at the origin) 6 Eample 4 : Let = + + 6 The -intercept is 6 since = 6 when = 0 To find the -intercepts, one must solve the equation 0 = + 6 0 = ( + 3)( ) This implies that we can have either ( + 3) = 0 which gives = 3 or ( ) = 0 which gives = The -intercepts are then 3 and The equation we are 16
currentl dealing with is called a quadratic, and the intercepts don t give enough information to plot the graph Since all quadratic functions are smmetrical, the turning point will alwas occur half-wa between the -intercepts The equation = + 6 is that of a parabola 6 4 4 4 6 The worksheet on differentiation gives us another method for finding the coordinates of A Point A is called a turning point because the graph changes direction at point A Another important output of functions that can be seen from graphs are the values of and where the function isn t defined These will appear as breaks in the graph For eample, we noted in the last worksheet that the function = 1 is not defined at = 0 Also, there is no input that will give an output of = 0, ie there are no or -intercepts The graph of = 1 looks like this: f() = 1 There is a break in the graph of this hperbola at = 0 and at = 0 17
Eample 5 : = g() 1 The function g() is not defined for 1 < < 0 Eercises: 1 Graph the following equations of straight lines b first finding and intercepts (a) = + (b) = 3 (c) = + 4 (d) + 6 = 0 (e) = 4 Section 3 Odd and Even Functions Some functions can be classed as odd or even functions Man functions, however, are not odd or even If we know that a certain function is odd or even, it will help us draw the graph An even function is one in which f(a) = f( a) for all a That is,, whatever number we choose as input, the output of the function will be the same if we change the sign of the input For an even function, if an input of c gives an output of b, then the input c also gives an output of b In function notation, this sas that if f( c) = b, then f(c) = b also Eample 1 : The function f() = is even f(1) = 1 f() = 4 f(a) = a f( 1) = 1 f( ) = 4 f( a) = a 18
An even function has a distinctive shape when graphed - the graph for the negative s (the left-hand side of the -ais) is a reflection of what is on the right-hand side of the -ais Each of the four graphs represents an even function Eample : Sa we are given part of a graph of g() We now etend the graph of g() to make an even function: 19
We now define what an odd function is A function f() is odd if f( ) = f() That is, if an input of a gives an output of b, then an input of a will give an output of b Eample 3 : The function g() = is an odd function Note that g(1) = 1 g() = g( 1) = 1 g( ) = The graph of an odd function has distinctive features; it has the propert that ou get what is on the right-hand side of the -ais b rotating what is on the left-hand side of the -ais through 180 And vice versa Eample 4 : = g() Here are more eamples of odd functions 0
Given the function h(): = h() We can etend h() to be an odd function: = h() If we are given a function as an epression, we can test it to see if it is odd or even b substituting in a and a as inputs and finding out what the outputs are Eample 5 : Is the function () = + 7 even, odd, or neither? (a) = a + 7 ( a) = ( a) + 7 = a + 7 We can see that ( a) = (a), so () is even Eercises: 1 Are the following functions even, odd, or neither? (a) f() = (b) f() = (c) f() = + 1 (d) f() = 3 1
(e) f() = + + 3 (f) (g) ( 1, ) (1, ) (, 3) (, 3) (h) Section 4 The equation of a Circle There are some equations that ou should be able to recognize at first glance, and know roughl what the look like: the equation of a straight line is an eample Another equation that ou should be able to recognize is the equation of a circle It is + = r for some number r When graphed, the set of points satisfing + = r will be a circle of radius r centered at the origin This means that the and -intercepts are ±r The equation of a circle is not actuall a function of since each value of has two possible values of in the domain r < < r But it is an equation that ou will be epected to know how to graph A variation of the equation of a circle alread given to ou is ( a) + ( b) = r This is still an equation of a circle, but it is more general: it has a radius r as before, but is centred on (a, b) rather than at the origin The circle has centre (, 1) and radius 5 The circle has centre ( 1, ) and radius ( ) + ( 1) = 5 ( + 1) + ( ) = 4
4 7 4 6 (, 1) ( ) + ( 1) = 5 1 1 4 ( + 1) + ( ) = 4 ( 1, ) Eercises: 1 Write down the equation of a circle which has (a) radius 3 and centre (6, ) (b) radius 5 and centre (4, 3) (c) radius 1 1 and centre ( 1, ) (d) radius and centre ( 1, 1 1 ) (e) radius 4 and centre ( 1, 3) 3
Eercises for Worksheet 3 1 (a) What is the range of the function whose graph is below? Give our answer in interval notation (b) What is the domain of the function whose graph is below (in interval notation)? 3 1 (c) What are the and intercepts of = 3? (d) The function = 3 4 crosses the -ais twice and the -ais once Find all three intercepts (e) Let f() = 3 i Show that f() is odd ii Sketch f() State whether the following equations represent a line, a parabola, a cubic, a circle, or a hperbola (a) + = 16 (b) = + 5 + 6 (c) = (d) ( 1) + ( + ) = 49 (e) = + 1 4
3 State whether the following functions are even, odd or neither: (a) (b) (c) 5
Worksheet 33 Trigonometr Section 1 Review of Trig Ratios Worksheet 8 introduces the trig ratios of sine, cosine, and tangent To review the ratios, consider a triangle ABC with angle φ as marked B c a φ A b C The hpotenuse (hp) of the triangle is c; the adjacent (adj) side is b; the opposite (opp) side is a The side of length a is opposite the angle, and the side of length b is the side adjacent to the angle which is not the hpotenuse Then we have sin φ = opp hp = a c cos φ = adj hp = b c tan φ = opp adj = a b Note also that a sin φ cos φ = c b c = a b = tan φ Eercises: 1 For the following triangle, find the ratios: (a) (b) (c) sin θ tan θ cos θ 3 5 θ 4 6
For the following triangle, find the ratios: (a) tan θ (d) sin φ (b) cos φ (e) tan φ (c) sin θ (f) cos θ a θ c b φ 3 (a) Use Pthagoras theorem to find (b) Find (i) sin θ (ii) tan θ (iii) cos θ 5 θ 1 Section Degrees and Radians Recall from Worksheet 9 that 1 = π 180 radians In universit maths it is much more common to give angles in radians rather than degrees If the units are left off an angle, then the angle is in radians Degrees can be converted to radians using the above formula, but it will be ver convenient for ou to know some standard conversions In particular: 90 = π 30 = π 6 45 = π 4 180 = π 60 = π 3 360 = π Eample 1 : An equilateral has three equal angles of π 3 Eample : Convert 50 to radians 50 = 50 π 180 = 5π 18 radians Eample 3 : How man degrees is π 9 radians? We know 180 = π, so π 9 = 180 9 therefore π 9 = 0 7
If we think of an angle θ as the amount of rotation of a straight line about the angle, then we can define a positive rotation as one which is anti-clockwise and a negative rotation as one which is in a clockwise direction We can see the ordinar plane with the verte of the angle at the origin and the base of the angle beginning at the positive -ais ie the positive -ais represents the position of the line if the angle of rotation is 0 So for the angle θ we get the following diagram: θ We can now represent angles graphicall and we can deal with angles of an size Recall that a full revolution is π, or 360 So rotating a line through π will bring it back to its starting position Eample 4 : Represent π radians graphicall Since π is the angle half wa round 3 the plane, π is the angle one third of the wa around the upper half of the plane 3 π 3 Eample 5 : Represent π 3 radians graphicall π 3 Eample 6 : Represent π 4 radians graphicall π 4 8
Eercises: 1 Write the following degrees in radian measure (a) 80 (b) 10 (c) 90 (d) 70 (e) 135 Convert the following radian measures to degrees (a) π 8 (b) π 9 (c) 3π 4 (d) 5π 6 (e) 7π 6 Section 3 Standard Triangles There are two triangles which are known as standard triangles These triangles and the information contained in them should be memorized, as ou will be epected to know certain information without using a calculator The first triangle is a right-angled isosceles triangle Recall that an isosceles triangle has two angles the same and two sides the same length: π 4 1 The associated trig ratios are fairl simple to work out, and are left as eercises The second standard triangle is half an equilateral triangle of length π π 3 3 When we have taken half the equilateral triangle, we end up with the following: π 3 π 6 3 π 3 1 Pthagoras theorem gives us the length of the vertical side as 3, and the angle φ is half the top angle so φ = π The trig ratios given b this triangle are: 6 9 1
sin π 3 = 3 sin π 6 = 1 cos π 3 = 1 cos π 6 = 3 tan π 3 = 3 tan π 6 = 1 3 Once the triangles are memorized the trig ratios can be found b just drawing either of the two triangles It is important that ou memorize the trig ratios for the angles π 6, π 3 and π 4 Eercises: 1 Find the eact ratios for the following (a) tan π 3 (b) cos π 6 (c) sin π 4 (d) tan π 4 (e) tan π 6 Use eact ratios to find θ in each of the following equations, where 0 θ π (a) sin θ = 1 (b) cos θ = 1 (c) tan θ = 1 (d) sin θ = 3 (e) cos θ = 1 Section 4 Using trigonometric ratios We can use the trigonometric ratios described in the previous sections to find an unknown angle or side in a right-angled triangle Consider the following triangle: θ 6 8 Let us sa that we wish to find θ in this triangle The side that is 6 units long is adjacent to θ; the side that is 8 units long is opposite to θ, so we have tan θ = opposite adjacent = 8 6 = 4 3 30
Consequentl, θ is an angle whose tangent is 4 That is θ = 3 tan 1 4 B using the tan 1 3 button on a calculator, we find that θ = 097, to three decimal places Note that this answer is in radians Eample 1 : What is θ? θ 0 7 With respect to θ, the opposite side is 7 units long, and the hpotenuse is 0 units long Therefore, sin θ = θ opposite hpotenuse = 7 0 = 0358 The last step was carried out using the sin 1 button on a calculator, and the answer is approimate and in radians Eample : What is? 10 06 The trigonometric ratios ma also be used to find the length of a side in a right angled triangle sin 06 = OPP HYP = 10 = 10 sin 06 = 581 Eample 3 : What is? 5 041 31
tan 041 = OPP ADJ = 5 = 5 tan 041 = 1087 Eercises: 1 Find θ in each of the following 4 θ 8 (a) θ (b) 5 6 5 (c) θ 1 (d) 4 θ 17 Find, to decimal places, in each of the following triangles 06 1 8 0 π 8 (a) (b) (c) 3 (d) π 8 54 Section 5 Inverse Trig Functions Sometimes ou will come across the notation sin 1 a or cos 1 a Now, sin 1 a does not mean 1 It is called the arcsine of a, and means this: the sin of what angle will give an answer a? sin a So sin 1 a = θ means sin θ = a The same rule applies to cos 1 a and tan 1 1 a If ou wish to write as (sin a) 1 so that there is no confusion sin a then ou would do so 3
1 (a) Covert i 5π 6 to degrees Eercises for Worksheet 33 ii 1π 9 to degrees iii 80 to radians; write the answer as a number times π iv 4 to radians; write the answer to 3 decimal places (b) Find the eact values of i sin π 4 ii cos π 6 iii tan π 3 1 (a) Find the value of 6 θ (b) Evaluate θ 3 Joan walks 5km north, then 36km east (a) Put these distances onto the appropriate sides of the triangle below: θ (b) Find the angle θ, the bearing that Joan has effectivel walked along 33
Worksheet 34 Further Trigonometr Section 1 Trig ratios for angles of an magnitude Recall from the last worksheet how we described a wa of drawing angles of an magnitude on the cartesian plane If we use the positive -ais to represent our starting point, then rotate this ais in an anticlockwise direction through α radians, we have an angle of α radians (with α positive) A negative angle can be drawn b rotation in a clockwise direction Eample 1 : We draw the angles π 6 and π 3 π 6 π 3 Instead of representing a point P using and coordinates, we could represent it as an angle of rotation and a distance awa from the origin P d θ As eamples consider two points: Q, which is units awa from the origin and rotated through an angle of 5π 4 ; R, which is 15 units awa from the origin and rotated through an angle of π 4 Q 5π 4 π 4 15 R The question now is how to connect this method of specifing a point with the usual wa of using the and coordinates In a previous section we defined sin θ and cos θ using right angled triangles A much more useful definition is the following Let a point P be eactl one 34
unit awa from the origin, and rotated b an angle θ Then the coordinate of P is defined to be cos θ, and the coordinate is defined to be sin θ The illustration of the definition is: P (cos θ, sin θ) 1 θ Eample : Draw a picture to determine whether sin( π ) and cos( π ) are positive 4 4 or negative π 4 1 P ( 1, 1 ) From the definitions of sine and cosine we have 1 = cos( π 4 ) 1 = sin( π 4 ) It is apparent from the picture that cos( π) > 0 and sin( π) < 0 4 4 Eample 3 : Draw a picture to locate the point (sin( 7π 6 ), cos( 7π 6 )) 7π 6 1 (cos 7π 6, sin 7π 6 ) Notice what happens when we appl this definition to an angle that is between 0 and π Let Q be the point that is eactl one unit awa from the origin, and rotated b an angle φ, where 35
0 < φ < π Sa the coordinate of Q is b and the coordinate is a The relevant picture is: Q a 1 φ b If ou think about the right angled triangle formed be the points (0, 0), (b, 0), and (a, b) and appl the right angled definitions of sine and cosine then ou get the formulae sin φ = a 1 = a cos φ = b 1 = b which is to sa the and coordinates of the point Q are cos φ and sin φ respectivel This is eactl the definition that we have just proposed! The point is that the definitions of sine and cosine that we have seen before, in terms of right angled triangles, match the new definition that we have just given in the case that the angles are between 0 and π (0 and 90 ) The advantage with the new definition is that it allows us to find the sine and cosine of angles of an magnitude, as well as for negative angles We do this b drawing the unit circle (which is a circle of radius 1 centred on the origin) An point on the circle is then eactl one unit awa from the origin Now, drawing in our angle from eample, π, we get 4 π 4 (1, 0) ( 1, 1 ) Now, 1 = cos( π 4 ) 1 = sin( π 4 ) 36
Notice that the triangle defined b the points (0, 0), ( 1, 0) and ( 1, 1 ) is a right angled triangle; the hpotenuse is of length 1 because the the radius of the unit circle is of length 1 It is drawn here: π 4 1 What remains is to find the length of the two sides of the triangle, which we will do b recalling our standard triangles from an earlier worksheet and using the properties of similar triangles The two standard triangles we looked at were 1 π 4 1 1 We can make the triangle on the left the same as the one that we took out of the unit circle b dividing all the lengths b a factor of Similarl, we can make the standard triangle on the right have a hpotenuse of length 1 b dividing each side b π 3 3 π 4 1 1 1 1 π 3 1 3 Putting the lengths back onto the unit circle picture gives us 1 π 4 1 (1, 0) ( 1, 1 ) Then we have 1 = cos( π 4 ) = 1 1 = sin( π 4 ) = 1 37
Eample 4 : Calculate the sine and cosine of 7π 6 (1, 0) 7π 6 (, ) Recalling the definitions of sine and cosine, we have = cos( 7π 6 ) = sin( 7π 6 ) B etracting the right angled triangle which connects the points (0, 0), (, 0), and (, ) and comparing it to the scaled standard triangles, we can put the following distances onto the unit circle diagram (1, 0) 7π 6 (, ) 1 3 As a result, we get = cos( 7π 6 ) = 3 = sin( 7π 6 ) = 1 What happens if an angle is bigger than π or less than π? Since a full revolution of a circle is π radians, the position on the circle is unchanged if we go an angle of θ or an angle of 38
θ + π If the position on the circle is unchanged b adding an angle of π, and the sines and cosines are defined in terms of the coordinates of appropriate points on the circle, then the sines and cosines of angles are unchanged b adding or subtracting multiples of π radians As an eample, graph the angle 11π 4 on the cartesian plane 11π 4 Notice that we would end up with an angle pointing in the same direction if we had performed a rotation of 11π 4 π = 3π 4 Eercises: 1 Find the eact ratio for each of the following (a) sin π 4 (b) tan( π 4 ) (c) cos π 6 (d) cos 7π 6 (e) sin( π 3 ) (f) tan 3π 4 (g) cos 3π 4 (h) sin( π 3 ) (i) tan 5π 6 (j) cos( π 3 ) Use a calculator to find the following to decimal places (a) cos 16 (b) sin π 8 (c) tan 7π 6 (d) cos( π 7 ) (e) sin( 06) (f) sin π 9 (g) cos 8π 5 (h) tan( π 9 ) 39
Section Graphs of Trig Functions The trig functions can be graphed on a Cartesian plane as functions of The unit of measurement for is radians It is helpful to be able to recognize the graphs of the main trig functions The function = sin is odd, with an -intercept ever integer multiple of π It looks like this: 1 π 3π π π 05 05 1 π π 3π π The function = sin is also periodic, with period π This means that sin() = sin( + π) for all values of If a function f() is periodic with period b, then f() = f( + b) for all We can see from the graph of = sin that the range of the function is [ 1, 1] ie 1 sin 1 for all The function = cos is even It is also periodic with period π The -intercept is 1 and the -intercepts are at π + kπ for integer k The graph of = cos looks like this: π π π 1 05 05 1 π π π Notice that the range is also [ 1, 1], so 1 cos 1 for all The graphs of sin and cos will help ou to remember the values of sin and cos for = 0, π, π, 3π and π if ou are a visual person Some people find it easier to remember the pictures, other people the numbers From the graph we get 40
sin( π) = 0 cos( π) = 1 sin( π ) = 1 cos( π ) = 0 sin( π ) = 1 cos( π ) = 0 sin( 3π ) = 1 cos( 3π ) = 0 sin(π) = 0 cos(π) = 1 The graph of = tan looks completel different from either cos or sin It is a periodic function with period π and it looks like this: π 3π π π π π 3π π Notice that the -intercepts are integer multiples of π, and that the -intercept is 0 Notice also that = tan is not defined at π + kπ for an integer k Recall that tan = sin cos so tan is undefined when cos = 0, which is at π + kπ for an integer k Eercises: 1 (a) Given = sin, complete the table of values π 5π π π 0 6 6 π 6 π 5π 6 π (b) Using the table draw the graph of = sin for π π 41
Section 3 Pthagorean Identities There are some equalities known as trigonometric identities which are ver useful in solving some kinds of problems The first one that we look at is derived from Pthagoras theorem Recall: From the above relations, we then have: sin θ = a/c cos θ = b/c a c a + b = c θ b sin θ + cos θ = ( a c ) + ( b c ) Then we have that for an angle θ: = a c + b c = a + b c = c c = 1 sin θ + cos θ = 1 The net two identities are also important, but will not be derived For an angles A and B: sin(a + B) = sin A cos B + sin B cos A cos(a + B) = cos A cos B sin A sin B These identities can be used to find the cos and sin of an angles bigger than π We can derive more trig identities from the ones that we alread have Eample 1 : sin = sin( + ) = sin cos + sin cos = sin cos 4
Eample : cos = cos( + ) = cos cos sin sin = cos sin Recall that = cos is an even function, therefore cos( ) = cos() Recall that = sin is an odd function, therefore sin( ) = sin() Eample 3 : sin(a B) = sin A cos( B) + sin( B) cos A = sin A cos B sin B cos A Eample 4 : cos(a B) = cos A cos( B) sin( B) sin A = cos A cos B + sin B sin A These identities can be used in man was One use for them is an alternative wa of finding the trig ratios of angles between 0 and π Eample 5 : Calculate the sine and cosine of 7π 6 sin 7π 6 = sin(π + π 6 ) = sin π cos π 6 + cos π sin π 6 3 = 0 + ( 1) 1 = 1 cos 7π 6 = cos(π + π 6 ) = cos π cos π 6 sin π sin π 6 = ( 1) 3 = 3 + (0) 1 43
Writing the angle 7π as π + π wasn t the onl option we could have used 7π = 6 6 6 3π π (Notice that the answers that we have here agree with the values calculated 3 using the unit circle earlier in the worksheet) Eercises: 1 Use eact ratios to show that sin π 6 + cos π 6 = 1 Use eact values to show that equation is satisfied when A = 0 and B = π 3 sin(a + B) = sin A cos B + sin B cos A Section 4 Solving trigonometric equations In the previous section θ was given and we evaluated the trigonometric ratios for the angle Now we investigate the situataion where we must find the value, or values, of θ when we are given a trigonometric ratio Eample 1 : Solve sin θ = 1 for 0 θ π Recall that sin θ is the coordinate of a point on the unit circle The first step then will be to draw a unit circle and draw the line = 1 ; the net, and last, step is to determine the angles of the points on the unit circle where the line = 1 cuts We draw two pictures, one with the basic information we have just outlined, and one with a few distances that have been worked out = 1 (1, 0) 1 = 1 1 1 1 3 3 (1, 0) 44
The first thing to note is that there are two solutions The lengths shown have been figured out using the fact that the vertical distances are 1, the fact that the radius of the circle is 1, and b recognizing that the triangles hidden in the picture are scaled versions of the standard triangles (which are shown in section 1) Given that we know the angles in the standard triangles, we can read off the angles to the two solutions as π and π π = 5π 6 6 6 Eample : Solve cos θ = 1 4 This must have solutions given b cos θ = 1 and cos θ = 1 The definition of cos θ is that it is the coordinate of the point defined b some angle around the unit circle So the solution will be obtained b drawing the lines = 1 and = 1, locating the points of intersection with the unit circle, then finding the appropriate angles Again we draw two pictures: one with minimal information, so we can see roughl where the solutions are as well as how man solutions there are; the other picture has details of distances and so on = 1 = 1 (1, 0) 1 1 3 (1, 0) From the first picture we can see that there are four solutions, as well as the fact that there is one solution in each quadrant Of the four triangles in the second picture, onl the distances for one of them have been shown, as all the other triangles are similar The distances have again been found b scaling a standard triangle from section 1 As the angles of the standard triangle are known, so are the angles of the four points shown The are π, π π, π + π, and π π (Another wa to 3 3 3 3 write the solutions would be π, π π, π, and (π π)) 3 3 3 3 45
Eercises: 1 Solve the following equations for θ; restrict our answers to 0 θ π (a) sin θ = 3 (b) tan θ = 3 (c) cos θ = 1 (d) cos θ = 3 (e) tan θ = 1 (f) sin θ = 1 46
1 Find the eact ratios of Eercises for Worksheet 34 (a) sin 3π 4 (b) tan π 6 (c) cos 7π 4 (d) cos 4π 3 (e) sin 5π 6 (f) tan π 3 (a) Use the epansion sin(a + B) = sin A cos B + sin B cos A to find the eact value of sin 7π 7π Note that = π + π 1 1 3 4 (b) Use the epansion cos(a B) = cos A cos B + sin A sin B to find the eact value of cos π Note that π = π π 1 1 3 4 3 Solve for θ in the interval 0 θ π (a) tan θ = 3 (b) sin θ = 1 (c) cos θ = 1 47
Worksheet 35 Simultaneous Equations Section 1 Number of Solutions to Simultaneous Equations In maths we are sometimes confronted with two equations in two variables and we want to find out which values of the variables satisf both of the equations Sometimes there will be no values of the variables that allow both equations to hold, and other equations will have man possible values of the variables The process of finding solutions is called solving simultaneous equations For eample, we might be asked to find the and values that allow both of the following equations to be true: 5 + = 3 4 + = 4 Recall from worksheet 10 that the general equation of a line can be written in the form a + b + c = 0, for a, b, and c constants Both of the equations above have the correct form for the equation of a line This is alwas the case when solving linear simultaneous equations in two variables This means that solving simultaneous equations is the same as finding the point of intersection of lines If certain values of and satisf both equations, the point (, ) will lie on both the lines If we think about a sstem of simultaneous equations as representing lines on the cartesian plane we can tell how man solutions there will be to the equations without actuall solving them When we draw two lines on the plane, there are three possibilities: 1 The lines cross just once The lines never cross 3 The lines lie on top of each other The first case corresponds to a unique solution, ie there is onl one value for each variable that will satisf both equations The second case occurs when the two lines are parallel, and aren t touching Parallel lines have the same slope (gradient) For instance, the two lines 1 = + and = are parallel, and don t touch The have the same gradient, in this case, but the -intercepts are different Consequentl, the two lines never touch each other Equation 1 has a -intercept of, and equation has a -intercept of If we draw these lines, we get 48
1 1 The third case arises when the two equations represent the same line on the plane, and so touch everwhere The two equations 1 = 3 + 6 and = 6 + 1 lie on top of one another when graphed because if we take equation and divide both sides b, the we get equation 1 eactl If two lines lie on top of one another there are an infinite number of (, ) pairs that will satisf both equations Namel, ever pair (, ) that satisfies equation 1 will also satisf equation To check on the number of solutions to a sstem of simultaneous equations we can rearrange both equations to the slope-intercept form and then compare gradients and intercepts If the gradients are different, we will have a single (ie unique) solution If the gradients are the same, but the -intercepts are different, then we will have no solutions If the gradients are the same, and the -intercepts are the same, then there will be an infinite number of solutions Putting this algebraicall, if we have: Then = m 1 + b 1 = m + b If m 1 m then there is one solution If m 1 = m and b 1 b there are no solutions If m 1 = m and b 1 = b there is an infinite number of solutions 49
Eample 1 : How man solutions do the following simultaneous equations have? 1 3 + 6 = 9 + 10 = 4 Rearranging, we get = + 3 for equation 1 = 5 + for equation The gradients of the two lines are different, so there will be one solution Eample : How man solutions do the following simultaneous equations have? 1 5 + 10 = 5 + = Rearranging, we get = + 1 for equation 1 = + for equation The gradients of the two lines are the same, but the intercepts are different Then the lines are parallel, but don t touch There are no solutions to the sstem Eample 3 : How man solutions do the following simultaneous equations have? 1 5 + 10 = 4 0 = 10 Rearranging, we get = 5 + 5 for equation 1 = 5 + 5 for equation The gradients of the two lines are the same, and the intercepts are also the same Then the lines are on top of each other, and there are infinitel man solutions 50
Eercises: 1 How man solutions would each of the following pairs of equations have? (a) = + 1 = 3 (b) = 6 4 = 3 (c) + = 0 + 1 = 0 Check our answers b graphing the pairs of lines on a number plane Section Solving simultaneous equations The previous section discussed how man solutions there are to a sstem of simultaneous equations in unknowns (which we have been writing as and ) We will learn how to find solutions to a sstem of simultaneous equations b eample Given two equations and two unknowns, our objective is to reduce this to one equation and one unknown, which we know we can solve We can solve two equations simultaneousl b graphing them and finding their point of intersection Let us solve for the following sstem graphicall: (i) + 4 = 0 (ii) + = 0 Drawing the two lines on one graph, we get: 4 + = 0 4 + 4 = 0 51
From the graph, we can see that the point of intersection is (1, 3) Substituting = 1 and = 3 into the two equations, we see: (i) 1 + 3 4 = 0 (true) (ii) 1 3 + = 0 (true) Hence the point (1, 3) satisfies both equations Sometimes the point of intersection is not eas to read off the graph, so solving a sstem of equations algebraicall is often easier and more precise Eample 1 : Solve the sstem: (i) + = 4 (ii) = If we add equation (i) to equation (ii) the s will cancel: (i) + (ii) = = 1 We now substitute = 1 into either equation (i) or (ii) Let us choose equation (i) Then 1 + = 3 = 3 So the solution is = 1 and = 3 (You can, and should, check the solution b substituting the values of and into both equations (i) and (ii)) Eample : Solve the sstem (i) 3 + = 5 (ii) + = 3 There are the same number of s i equation (i) and (ii), so if we subtract the equations we would eliminate the s (i) - (ii) = 8 (Note 5 ( 3) = 8) = 4 5
Substitute = 4 in equation (ii): 4 + = 3 = 7 = 3 1 The solution is = 4, = 3 1 ; check that the solution satisfies (i) and (ii) Eample 3 : Solve the sstem Subtracting () from (1) gives (1) = + 3 () = + 5 (3) 0 = This is nonsense, and a check shows that there is no solution to this sstem because the lines that the equations represent are parallel Eample 4 : Solve the sstem (1) + 3 = 10 () 5 + 4 = 11 Sometimes it is necessar to multipl both equations b different numbers to get the same multiple of one of the variables Here is another eample of the usefulness of being able to find the lowest common multiple of two numbers We will take equation (1) times 5, and equation () times : Now subtracting (4) from (3): (3) 10 + 15 = 50 (4) 10 + 8 = 10 + 15 (10 + 8) = 50 7 = 8 which gives = 4 Substituting this back into equation (1) gives = 1 A check reveals that = 1 and = 4 is indeed a solution to the original equations 53
Eercises: 1 Solve graphicall the sstem of equations + = 3 = 1 Solve the following sstems algebraicall where possible (a) = 5 3 + = 10 (b) + 3 = 1 + = 4 (c) 3 + = 4 + = 5 (d) + 5 = 4 3 + = 6 54
Eercises for Worksheet 35 1 How man solutions (one, none, or infinite) will each of the following pairs of equations have? (a) (b) (c) (d) (e) + 3 = 5 + 3 = 9 + = 3 = 3 = 1 + 4 = + 5 = 3 4 + = = 1 3 + 9 3 = Solve the following sstems of equations (a) (b) (c) = 3 = + 5 = 4 + = 3 3 = 1 3 + 4 = 1 3 The sum of Peter and Anneka s ages is 4, and the difference between their ages is 6 Find their ages given that Peter is older than Anneka 55
Worksheet 36 Arithmetic and Geometric Progressions Section 1 Arithmetic Progression An arithmetic progression is a list of numbers where the difference between successive numbers is constant The terms in an arithmetic progression are usuall denoted as u 1, u, u 3 etc where u 1 is the initial term in the progression, u is the second term, and so on; u n is the nth term An eample of an arithmetic progression is, 4, 6, 8, 10, 1, 14, Since the difference between successive terms is constant, we have u 3 u = u u 1 and in general u n+1 u n = u u 1 We will denote the difference u u 1 as d, which is a common notation Eample 1 : Given that 3,7 and 11 are the first three terms in an arithmetic progression, what is d? 7 3 = 11 7 = 4 Then d = 4 That is, the common difference between the terms is 4 If we know the first term in an arithmetic progression, and the difference between terms, then we can work out the nth term, ie we can work out what an term will be The formula which tells us what the nth term in an arithmetic progression is where a is the first term u n = a + (n 1) d Eample : If the first 3 terms in an arithmetic progression are 3,7,11 then what is the 10th term? The first term is a = 3, and the common difference is d = 4 u n = a + (n 1)d u 10 = 3 + (10 1)4 = 3 + 9 4 = 39 56
Eample 3 : If the first 3 terms in an arithmetic progression are 8,5, then what is the 16th term? In this progression a = 8 and d = 3 u n = a + (n 1)d u 16 = 8 + (10 1) ( 3) = 37 Eample 4 : Given that, 5 and 6 are the first three terms in an arithmetic progression, what is d? 5 = (6 ) 5 = 4 Since = 4, the terms are 8, 5, and the difference is 3 The net term in the arithmetic progression will be 1 An arithmetic series is an arithmetic progression with plus signs between the terms instead of commas We can find the sum of the first n terms, which we will denote b S n, using another formula: S n = n [a + (n 1)d] Eample 5 : If the first 3 terms in an arithmetic progression are 3,7,11 then what is the sum of the first 10 terms? Note that a = 3, d = 4 and n = 10 S 10 = 10 ( 3 + (10 1) 4) = 5(6 + 36) = 10 Alternativel, but more tediousl, we add the first 10 terms together: S 10 = 3 + 7 + 11 + 15 + 19 + 3 + 7 + 31 + 35 + 39 = 10 This method would have drawbacks if we had to add 100 terms together! Eample 6 : If the first 3 terms in an arithmetic progression are 8,5, then what is the sum of the first 16 terms? S 16 = 16 ( 8 + (16 1) ( 3)) = 8(16 45) = 3 57
Eercises: 1 For each of the following arithmetic progressions, find the values of a, d, and the u n indicated (a) 1, 4, 7,, (u 10 ) (b) 8, 6, 4,, (u 1 ) (c) 8, 4, 0,, (u 0 ) (d) 0, 15, 10,, (u 6 ) (e) 40, 30, 0,, (u 18 ) (f) 6, 8, 10,, (u 1 ) (g), 1, 3,, (u 19) (h) 6, 5 3, 5 1,, (u 4 10) (i) 7, 6 1, 6,, (u 14) (j) 0, 5, 10,, (u 15 ) For each of the following arithmetic progressions, find the values of a, d, and the S n indicated (a) 1, 3, 5,, (S 8 ) (b), 5, 8,, (S 10 ) (c) 10, 7, 4,, (S 0 ) (d) 6, 6 1, 7,, (S 8) (e) 8, 7, 6,, (S 14 ) (f), 0,,, (S 5 ) (g) 0, 16, 1,, (S 4 ) (h) 40, 35, 30,, (S 11 ) (i) 1, 10 1, 9,, (S 9) (j) 8, 5,,, (S 0 ) Section Geometric Progressions A geometric progression is a list of terms as in an arithmetic progression but in this case the ratio of successive terms is a constant In other words, each term is a constant times the term that immediatel precedes it Let s write the terms in a geometric progression as u 1, u, u 3, u 4 and so on An eample of a geometric progression is 10, 100, 1000, 10000, Since the ratio of successive terms is constant, we have u 3 = u u u 1 u n+1 = u u n u 1 and The ratio of successive terms is usuall denoted b r and the first term again is usuall written a 58
Eample 1 : Find r for the geometric progression whose first three terms are, 4, 8 4 = 8 4 = Then r = Eample : Find r for the geometric progression whose first three terms are 5, 1, and 1 0 1 5 = 1 0 1 = 1 10 Then r = 1 10 If we know the first term in a geometric progression and the ratio between successive terms, then we can work out the value of an term in the geometric progression The nth term is given b u n = ar n 1 Again, a is the first term and r is the ratio Remember that ar n 1 (ar) n 1 Eample 3 : Given the first two terms in a geometric progression as and 4, what is the 10th term? a = r = 4 = Then u 10 = 9 = 104 Eample 4 : Given the first two terms in a geometric progression as 5 and 1, what is the 7th term? a = 5 r = 1 10 Then u 7 = 5 ( 1 10 )7 1 5 = 1000000 = 0000005 59
A geometric series is a geometric progression with plus signs between the terms instead of commas So an eample of a geometric series is 1 + 1 10 + 1 100 + 1 1000 + We can take the sum of the first n terms of a geometric series and this is denoted b S n : S n = a(1 rn ) 1 r Eample 5 : Given the first two terms of a geometric progression as and 4, what is the sum of the first 10 terms? We know that a = and r = Then S 10 = (1 10 ) 1 = 046 Eample 6 : Given the first two terms of a geometric progression as 5 and 1, what is the sum of the first 7 terms? We know that a = 5 and r = 1 10 Then S 7 = 5(1 1 7 ) 10 1 1 10 = 5 1 1 10 7 9 10 = 5555555 In certain cases, the sum of the terms in a geometric progression has a limit (note that this is summing together an infinite number of terms) A series like this has a limit partl because each successive term we are adding is smaller and smaller (but this fact in itself is not enough to sa that the limiting sum eists) When the sum of a geometric series has a limit we sa that S eists and we can find the limit of the sum For more information on limits, see worksheet 37 The condition that S eists is that r is greater than 1 but less than 1, ie r < 1 If this is the case, then we can use the formula for S n above and let n grow arbitraril big so that r n becomes as close as we like to zero Then S = a 1 r is the limit of the geometric progression so long as 1 < r < 1 60